Question

a 0.25 kg arrow with a velocity of 15m/s to the east strikes and pierces the bull eye of a 7.0 kg target

Answer 1

Given: mA = 0.02 kgvA1= 0.25 m/svA2 = -0.16 m/smB = 0.04 kgvB1 = -0.15 m/sFind: vB2 (velocity of Ball B after collsion)Solution:vB2 = mAvA1 + mBvB1 – mAvA2mB= (0.02 kg)(0.25 m/s) + (0.04 kg)(-0.15 m/s) – (0.02 kg)(-0.16 m/s)0.04 kg = 0.0022 kg.m/s0.04 kg= 0.055 m/s